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Let $\frak{g}$ be a semisimple Lie algebra, and let $({-},{-})$ be an invariant inner product on $\frak{g}$. The Chevalley–Eilenberg complex $C^*(\frak{g})$ has a natural Poisson bracket of degree $-2$; if we think of the generators of the exterior algebra $C^*(\frak{g})$ as coordinates $\xi$ on a graded manifold, then this Poisson bracket is associated to the symplectic form $\Omega=(d\xi,d\xi)$.

The resulting (shifted) differential graded Lie algebra is formal, and the Poisson bracket induces the vanishing bracket on the cohomology. I have written down a proof, which uses explicit generators for the cohomology (essentially, the Chern–Simons classes). Is there a published proof of this result, perhaps less computational?

After posting this question, I learned that this construction is discussed in the article

Shifted Poisson and symplectic structures on derived N-stacks Jon Pridham (Examples 3.31)

Unless I am mistaken, the special case where $\mathfrak{g}$ is semisimple is not addressed in Pridham's article. I am grateful for all of the general bibliographic references, but I am interested in a very specific result, and none of the comments below address the question I asked.

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    $\begingroup$ Something doesn’t smell right. What if g is abelian? Then the differential is zero, so it is automatically formal, but the bracket is not zero. $\endgroup$ Oct 13 at 21:24
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    $\begingroup$ Those co-ordinates are in the wrong degree to give a derived affine scheme. That's a form of Lie algebroid, enhancing things in the opposite direction, so $H^*$-isomorphism is too weak to give an equivalence on the associated stacks. $\endgroup$ Oct 13 at 21:27
  • $\begingroup$ Isn’t this the standard 2-shifted symplectic structure on BG,(associated to an invariant inner product) pulled back to its formal completion (B exp(g) )? $\endgroup$ Oct 13 at 22:58
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    $\begingroup$ @DavidBen-Zvi : yes, as in Examples 3.31 of arxiv.org/abs/1504.01940 , for instance. $\endgroup$ Oct 13 at 23:14
  • $\begingroup$ There was a misprint in the original question, which Theo has pointed out: I meant to write "semisimple", not "reductive", on the first line. $\endgroup$ 2 days ago

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